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stiks02 [169]
1 year ago
12

Which is the least likely cause of an engine to hunt and surge at top no-load speeds? A lean air/fuel mixture An incorrect spark

plug heat range. A blocked carburetor bowl vent All of the above.
Physics
2 answers:
77julia77 [94]1 year ago
8 0
I would say that the answer to this is the last option: ALL OF THE ABOVE. A lean fuel or air mixture, incorrect spark plug heat range, and a blocked carburetor bowl vent would be among the reasons of an engine to surge and hunt at the top no-load speeds. Hope this answer helps.
Zolol [24]1 year ago
4 0

Answer:

A blocked  carburetor bowl vent

Explanation:

Engine hunt is defined as the continuous  variation in rpm of the engine. It can occur due to various conditions and abnormalities in the inside the fuel injector pump.

In ideal situation, amount of fuel given to the injector is fixed  but in some cases such abnormality arises hence rpm get changed.

Hence of the options available, a blocked carburetor bowl is the least likely cause for engine hunt.

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13. An aircraft heads North at 320 km/h rel:
AURORKA [14]

The velocity of the aircraft relative to the ground is 240 km/h North

Explanation:

We can solve this problem by using vector addition. In fact, the velocity of the aircraft relative to the ground is the (vector) sum between the velocity of the aircraft relative to the air and the velocity of the air relative to the ground.

Mathematically:

v' = v + v_a

where

v' is the velocity of the aircraft relative to the ground

v is the velocity of the aircraft relative to the air

v_a is the velocity of the air relative to the ground.

Taking north as positive direction, we have:

v = +320 km/h

v_a = -80 km/h (since the air is moving from North)

Therefore, we find

v'=+320 + (-80) = +240 km/h (north)

Learn more about vector addition:

brainly.com/question/4945130

brainly.com/question/5892298

#LearnwithBrainly

7 0
2 years ago
A 50-kg load is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. By what distance will the wire stretch? Young'
lbvjy [14]

Answer:

3.5 cm

Explanation:

mass, m = 50 kg

diameter = 1 mm

radius, r = half of diameter = 0.5 mm = 0.5 x 10^-3 m

L = 11.2 m

Y = 2 x 10^11 Pa

Area of crossection of wire = π r² = 3.14 x 0.5 x 10^-3 x 0.5 x 10^-3  

                                              = 7.85 x 10^-7 m^2

Let the wire is stretch by ΔL.

The formula for Young's modulus is given by

Y =\frac{mgL}{A\Delta L}

\Delta L =\frac{mgL}{A\times Y}

ΔL = 0.035 m = 3.5 cm

Thus, the length of the wire stretch by 3.5 cm.

5 0
2 years ago
Wrapping paper is being unwrapped from a 5.0-cm radius tube, free to rotate on its axis. if it is pulled at the constant rate of
lisov135 [29]
So the equation for angular velocity is

Omega = 2(3.14)/T

Where T is the total period in which the cylinder completes one revolution.

In order to find T, the tangential velocity is

V = 2(3.14)r/T

When calculated, I got V = 3.14

When you enter that into the angular velocity equation, you should get 2m/s
5 0
2 years ago
A tin can whirled on the end of a string moves in a circle because
Ilya [14]

Answer:

There is an inward force acting on the can

Explanation:

This inward force is known as Centripetal force and it is responsible for making the can whirl on the end of a string in circle and it is also directed towards the center around which the can is moving.

8 0
2 years ago
An automobile traveling at 25.0 km/h along a straight, level road accelerates to 65.0 km/h in 6.00 s. what is the magnitude of t
USPshnik [31]
Note that
1 km/h = (1000 m)/(3600 s) = 0.27778 m/s

Initial velocity, v₁ = 25 km/h = 6.9444 m/s
Final velocity, v₂ = 65 km/h = 18.0556 m/s

Time interval, dt = 6 s.

Calculate average acceleration.
a = (v₂ - v₁)/dt
   = (18.0556 - 6.9444 m/s)/(6 s)
   = 1.852 m/s²

Answer:
The average acceleration is 1.85 m/s² (nearest hundredth)
3 0
2 years ago
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